Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 14"
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== Problem == | == Problem == | ||
− | In triangle ABC, <math> | + | In [[triangle]] <math>ABC</math>, <math>AB = 308</math> and <math>AC=35</math>. Given that <math>AD</math>, <math>BE,</math> and <math>CF,</math> [[intersect]] at <math>P</math> and are an [[angle bisector]], [[median of a triangle | median]], and [[altitude]] of the triangle, respectively, compute the [[length]] of <math>BC.</math> |
[[Image:Mock AIME 2 2007 Problem14.jpg]] | [[Image:Mock AIME 2 2007 Problem14.jpg]] | ||
+ | |||
+ | ==Solution== | ||
+ | Let <math>BC = x</math>. | ||
+ | |||
+ | By the [[Angle Bisector Theorem]], <math>\frac{CD}{BD} = \frac{AC}{AB} = \frac{35}{308} = \frac{5}{44}</math>. | ||
+ | |||
+ | Let <math>CF = h</math>. Then by the [[Pythagorean Theorem]], <math>h^2 + AF^2 = 35^2</math> and <math>h^2 + BF^2 = x^2</math>. Subtracting the former [[equation]] from the latter to eliminate <math>h</math>, we have <math>BF^2 - AF^2 = x^2 - 35^2</math> so <math>(BF + AF)(BF - AF) = x^2 - 1225</math>. Since <math>BF + AF = AB = 308</math>, <math>BF - AF = \frac{x^2 - 1225}{308}</math>. We can solve these equations for <math>BF</math> and <math>AF</math> in terms of <math>x</math> to find that <math>BF = 154 + \frac{x^2 - 1225}{616} = </math> and <math>AF = 154 - \frac{x^2 - 1225}{616}</math>. | ||
+ | |||
+ | Now, by [[Ceva's Theorem]], <math>\frac{AE}{EC} \cdot \frac{CD}{DB} \cdot \frac{BF}{FA} = 1</math>, so <math>1 \cdot \frac{5}{44} \cdot \frac{BF}{AF} = 1</math> and <math>5BF = 44AF</math>. Plugging in the values we previously found, | ||
+ | |||
+ | <math>5\cdot 154 + \frac{5(x^2 - 1225)}{616} = 44\cdot 154 - \frac{44(x^2 - 1225)}{616}</math> | ||
+ | |||
+ | so | ||
+ | |||
+ | <math>\frac{49}{616}(x^2 - 1225) = 39\cdot 154</math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>x^2 - 1225 = 75504</math> | ||
+ | |||
+ | which yields finally <math>x = 277</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{Mock AIME box|year=2006-2007|n=2|num-b=13|num-a=15}} | ||
+ | |||
== Problem Source == | == Problem Source == | ||
− | 4everwise thought of this problem after reading the first chapter of Geometry Revisited. | + | 4everwise thought of this problem after reading the first chapter of [http://www.amazon.com/exec/obidos/ASIN/0883856190/artofproblems-20 Geometry Revisited]. |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 10:53, 4 April 2012
Contents
Problem
In triangle , and . Given that , and intersect at and are an angle bisector, median, and altitude of the triangle, respectively, compute the length of
Solution
Let .
By the Angle Bisector Theorem, .
Let . Then by the Pythagorean Theorem, and . Subtracting the former equation from the latter to eliminate , we have so . Since , . We can solve these equations for and in terms of to find that and .
Now, by Ceva's Theorem, , so and . Plugging in the values we previously found,
so
and
which yields finally .
See Also
Mock AIME 2 2006-2007 (Problems, Source) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |
Problem Source
4everwise thought of this problem after reading the first chapter of Geometry Revisited.